# Shift operator (integral calculus involving Hermite polynomials) [closed]

I didn't know whether to pose this question on Physics.stackexchange or Math.stackexchange. But since this is the last step of a development involving the eigenfunctions of an Harmonic oscillator and a shift operator matrix, I thought it'd be better to post it here.

I have to calculate the integral

$$\frac{1}{2^nn!\sqrt{\pi}}\int_{-\infty}^{+\infty}H_n(x)e^{-x^2+kx}H_l(x)\;\mathrm{d}x$$

where $H_n(x)$ is the $n^{th}$ Hermite polynomial and prove that it equals

$$\sqrt{\frac{m_<!}{m_>!}}\left(\frac{k}{\sqrt{2}}\right)^{|n-l|}L_{m_<}^{|n-l|}\left(-\frac{k^2}{2}\right)\exp\left(\frac{k^2}{4}\right)$$

where $m_<$ and $m_>$ denote the smaller and the larger respectively of the two indices $n$ and $l$ and where $L_n^m$ are the associated Laguerre polynomials.

The last term is $\exp(k^2/4)$, hence I suppose that I begin with

$$\frac{1}{2^nn!\sqrt{\pi}}\int_{-\infty}^{+\infty}H_n(x)e^{-x^2+kx-\frac{k^2}{4}}e^{\frac{k^2}{4}}H_l(x)\;\mathrm{d}x$$ $$\frac{1}{2^nn!\sqrt{\pi}}e^{\frac{k^2}{4}}\int_{-\infty}^{+\infty}H_n(x)e^{-(x-\frac{k}{2})^2}H_l(x)\;\mathrm{d}x$$

but here I'm stuck... No matter what or how I can't go further.

• Cross-posted from math.stackexchange.com/q/299714/11127 Feb 15 '13 at 17:57
• Hi mwoua, and welcome to Physics Stack Exchange! It doesn't matter where this question came from; the fact is, it's about pure math, not physics, and that makes it off topic for us. Since you cross-posted it I'm closing it rather than migrating. (Please don't post the same question to more than one site in the future.) Feb 16 '13 at 0:38

One way to do this is by induction, first on $n$ and then on $l$. The base case is easy, since $H_0(x)$ is constant, and the integral is a simple gaussian; the integral for $n=1$ and $l=0$ is also easy. Then fix $l=1$ and assume the formula for arbitrary $n$. Then the formula can be proven for $n+1$ by using the recurrence relation for $H_{n+1}$, $$H_{n+1}(x)=2xH_n(x)-2nH_n(x),$$ changing the $2x$ factor for a derivative with respect to $k$, and applying a recurrence relation for the Laguerre polynomial on the right-hand side. That will prove the general case under $l=1$. Then using a similar induction procedure for $1\leq l\leq n$ will prove the full statement.
The other possibility is to do what everyone else does: reduce it to the matrix element $\langle m|\hat{D}(\alpha)|n\rangle$ and then blindly cite* Cahill and Glauber (Ordered expansions in boson amplitude operators. Phys. Rev. 177 no. 5 (1969), pp. 1857-1881, Appendix B.). What they do, if my thesis notes are to be trusted, is compare the matrix element $$\langle m|\hat{D}(\beta)|\alpha\rangle=\langle m|e^{\frac12 (\beta\alpha^\ast-\beta^\ast\alpha)}|\alpha+\beta\rangle=\frac{1}{\sqrt{m!}}(\beta+\alpha)^m e^{-\frac12|\beta|^2-\frac12|\alpha|^2-\beta^\ast\alpha}$$ to the generating function of the Laguerre polynomials, $$(1+y)^m e^{-xy}=\sum_{n=0}^\infty L_n^{(m-n)}(x) y^n$$ (which is valid for all $y\in\mathbb{C}$; take $y=\beta/\alpha$ up to conjugates) and from there to the original one expanding the coherent state $|\alpha\rangle$ in a number state expansion, comparing coefficients of $\alpha^n$.
(Note also that you will have to do a rotation to complex $k$. This is because your integral is of the form $\langle m|e^{k\hat{x}}|n\rangle$ and for real $k$ the operator $e^{k\hat{x}}$ is not unitary. Doing that also brings your desired result into the much nicer form $L_{m_<}^{|n-l|}(k^2/2)e^{-\frac14 k^2}$, which oscillates for small $k$ and then decays. Changing $k$ for $ik$ is valid because both sides of your target equality are entire functions of $k\in\mathbb{C}$, and proving them equal in one axis is enough by analytic continuation.)